How do I find the volume generated by revolving $(x - H)^2 + y^2 = a^2$ around the y axis specifically using cross sections (I know that shells are easy to use but I can't).
I have struggled with this problem for a long time, mainly because I can't find an $r_{\text{in}}$ and $r_{\text{out}}$. Please help (at least with those)!
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As a check, I use shells. They are really ribbons of radius $x$ and height $2 \sqrt{a^2-(x+H)^2}$ around the origin, of thickness $dx$. The volume of the donut is then
$$V = 2 \cdot 2 \pi \int_{H-a}^{H+a} dx \: x \, \sqrt{a^2-(x+H)^2}$$
This integral is easily done by a trig substitution of the form $x=H + a \sin{\theta}$. The integral then becomes
$$V = 4 \pi a^2 \int_{-\pi/2}^{\pi/2} d\theta \: (H + a \sin{\theta}) \cos^2{\theta}$$
This breaks into two pieces, and the second piece is easily shown to be zero. The volume is then
$$V = 4 \pi a^2 H \int_{-\pi/2}^{\pi/2} d\theta \: \cos^2{\theta} = 2 \pi^2 a^2 H$$
To find the volume by cross sections, note that you are simply revolving an area of $\pi a^2$ about a circumference of $2 \pi H$. The result is then
$$V = (\pi a^2) (2 \pi H ) = 2 \pi^2 a^2 H$$
Hint: consider integration by $y$ going from $-a$ to $a$. Draw a picture and you'll see that $$r_{in}=H-\sqrt{a^2-y^2}, r_{out}=H+\sqrt{a^2-y^2}$$
Edit: Check this and see where your error is. $$V=\pi\int_{-a}^{a}{(H+\sqrt{a^2-y^2})^2-(H-\sqrt{a^2-y^2})^2}dy=\pi\int_{-a}^{a}{4H\sqrt{a^2-y^2}}dy$$ Doing a substitution $y=a\sin t$, we will get. $$V=4\pi H\int_{y=-a}^{a}{a^2\cos^2 t}dt=2\pi H a^2 * (t-\frac{\sin 2t}{2})|_{y=-a}^a=2\pi^2 H a^2$$